Which element is oxidized in the following reaction: cuo(s) + h2 (g) → cu(s) + h2o (l)

Start by determining the oxidation numbers of each element on both sides of the equation CuO(s) + H2 (g) → Cu(s) + H2O (l)

Let's start with the reactants.

In the compound CuO(s), the oxidation number of O is -2 and the oxidation number of Cu is +2. (Remember, oxygen's oxidation number is -2 unless it's in a peroxide)

In H2(g) the oxidation number of H is 0 because it is in its elemental form.

Now onto the products

The oxidation number of Cu(s) is 0 because it is in its elemental form

In H2O(l) the oxidation number of oxygen is -2, and the oxidation number of H is +1.

Now let's summarize the change in the oxidation numbers of each element.

Cu: started as +2 and became 0

O: started as -2 and became -2 (didn't change)

H: started as 0 and became +1

To determine which element was oxidized, we look at which element lost electrons, and thus became more positive. So, we can determine that H was oxidized in this reaction, because it started with the oxidation number 0 and became +1.

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Explanation:

1 a

2 d

3 a

4 c

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